1.) Let \(a \neq 0\) and \(a^2 \equiv 0 \pmod{12}\). What is the value of a? Express your answer as a residue between 0 and the modulus.
2.) Find the last two digits of 18^37.
3.) Find the remainder when 100^100 is divided by 7.
1) a=6
2) 18^37 =27866516911699690485530919446962290200798035968
3)100^100 mod 7 =2
+118653 2.) Find the last two digits of 18^37.
\(18^1=18 \;(mod100)\\ 18^2=24 \;(mod100)\\ 18^3=32 \;(mod100)\\ 18^4=76 \;(mod100)\\ 18^5=68 \;(mod100)\\ 18^6=24 \;(mod100)\\ \mbox{and so the pattern continues}\\ 18^1=18 \;(mod100)\\ 18^2=24 \;(mod100)\\ 18^3=32 \;(mod100)\\ 18^{4N}=76 \;(mod100)\\ 18^{4N+1}=68 \;(mod100)\\ 18^{4N+2}=24 \;(mod100)\\ 18^{4N+3}=32 \;(mod100)\\ 18^{4N}=76 \;(mod100)\\ where \;\;N\in Z\;\;(integers)\;\;\;and\;\;\;N\ge1 \)
\(18^{37}=18^{4*9+1}=68\;\;mod\;100\)
So the last 2 digits will be 68 
